IBPS CLERK ::: TIME AND DISTANCE

in the concept of time, distance and speed is
related to a particular object in motion.

Relation between Time, Distance
and speed

Speed is the
distance covered by an object in one unit of time. It is calculated by dividing
the distance traveled by the time taken.

Speed
= Distance

Time

Distance = Speed X Time

Time
= Distance

Speed

Important points for time and
distance

·If speed is constant, then distance covered
by an object is proportional to time (more time, more distance).

·If Time is constant, then distance covered
by an object is proportional to its speed (More spped, more distance).

·If distance is constant, then speed is inversely
proportional to time. Therefore,

S1T1=S2T2=S3T3=……………

· a km\h
= (aX5/18) m/s.

· a m/s = (aX18/5) km/h

·If two objective are in motion and their
speeds are a and b, respectively, then

(a)Relative speed = a+b (if two objects are in
opposite direction)

(b)Relative speed = a-b (if two objects are in same direction)

·

If the ratio of speeds of A and B is x:y,
then the ratio oftime taken by them to cover the same distance is given by

1/x : 1/y, i.e., y: x

Example:

Convert 90 km/h
into m/s.

Answer:

We know that a km/h = (ax5/18) m/s

90 km/s = (90x5/18)m/s

5x5 = 25 m/s.

Example: A
train covers 135 km in 5 hh, find the speed of the train.

Answer: we know that

Speed = Distance / time.

Speed = 135/5 = 27 km/h.

Example :
a bike crosses a bridge with a speed of 108 km/h . what will be the length of
the bridge if the bike taken 8 min to cross the bridge?

Answer: here, the length of thew bridge

=
distance travelled by bike in 8 min

=
speed X time

Given that
speed is 108 km/h = 108X 5/18 m/s = 30 m\s

Time = 8 min = 8x 60 = 480 s

Lenth of
bridge = 480X 30 = 14400 m.

SHORT CUTS:

Technique 1:

When a
cetain distance is covered at a speed A and the same distance is covered at a
speed B, then the average speed during the whole journey is given by

= 2 AB /A+B

Example: mahesh covers a
certain distance by car driving at 35 km/h and he returns back to the starting
point riding on a bike with a speed of 25 km/h. find the average speed of the
whole journey.

Answer: Accroding to the formula

Average
speed = 2X35X25/60

= [A= 35 km/s, B= 25 km/s]

= 70X25/60 = 175/6

= 29.16
km/h.

Technique 2:

When
a person reaches a certain distance ‘a’ he gets t1 time late and when he
increases his speed by ‘b’ to cover the same distance, then he still gets late
by t2 time. In case, the distance is calcuated by

D = (t1
– t2)(a+b) a/b

Example: A
boy walking at a speed of 20 km/h reaches his school 30 min late. Next time he
increases his speed by 4 km/h but still he is late by 10 min. find the distance
of the school from his home?

Answer:Given data a=20 km/h, b= 4 km/h, t1= 30 min, t2
= 10 min

According to the formula,

Required
distance = (t1- t2)(a+b)a/b

=
(30-20)/60(20+4)20/4

=
20/60X24X20/4

=
5X8 = 40 km.

Technique 3:

When two persons A and B travel from points P
to Q, a distance of D with speeds ‘a’ and ‘b’, respectively and B reaches Q
first, returns immediately and meets A to R, then

Distance traveled by A (from points P to R) = 2X D(a/a+b)

Distance traveled by B (PQ+QR) = 2XD (b/a+b)

Example:
ramu and monu travel from Pto Q, a distance of 42 km, at 6 km/h and 8 km/h,
respectively. Monu reaches Q first and returns immediately and meets ramu at R.
find the distance from points P to R.

Answer: here, D=
42 kms, a= 6 km/h, b= 8 km/h

According
to the formula,

Distance
travelled by ramu = PR = 2Dxa/a+b

= 2X42X6/6+8

= 2X42X6/14

= 2X3X6 = 36 km

Technique 4:

A policeman sees a thief at a
distance of d. he starts chasing the thief who is running at a speed of ‘a’ and
policeman is chasing with a speed of ‘b’ (b>a). in this case, the distance
covered by the thief when he is caught by the policeman, is given by

d(a/b-a)

Example: A
policeman sees a chain snatcher at a distance of 50 m. He starts chasing the
chain snatcher who is running with a speed of 2 m/s while the policeman chasing
him with a speed of 4 m/s. find the distance covered by the chain snatcher when
he is caught by the policeman.

Answer: here, d= 50m, a= 2 m\s, b= 4m/s

According
to the formula, required distance = d(a/b-a)

= 50X 2/4-2

ans= 50 m

IBPS CLERK ::: TIME AND DISTANCE
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SSC-IBPS
on
14:09:00
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